An experiment that illustrates both chemical and electrolyticreduction by the pr

Question

An experiment that illustrates both chemical and electrolyticreduction by the preparation of ammonium vanadium (III) sulfate,(NH4)V(SO4)2.12H20 also known as vanadium alum. Ammonium metavanadate NH4VO3 is reduced by sulfur dioxide to vanadium IV as the vanadyl ion VO2+. This is then reduced by electrolysis ata graphite cathode to vanadium (III).

5g of ammonium metvanadate had 4ml of water added to it. Then20ml of SO2/sulfuric acid solution (1:4 concentrated sulfuric:water) was added which was saturated with SO2. An SO2 bubbler head was attached to the flask. The flask was stoppered and stirred until metavanadate was reduced to vanadylsulfate. The solution was then transferred to a beaker and boiled to expel excess sulfur dioxide.

An electrolytic reduction to vanadium (III) then took place with the vanadyl sulfate solution inside the porous pot and sulfuricacid outside.

Questions are:

What is the coordination geometry about the vanadium ion incrystals of the alum?

Write balanced equations for each reduction step

Assuming an average current of 1.75amps is passed during theelectrolysis, calculate how long it would take to reduce all of theV(IV) in the sample to V(III)?

In this experiment reduction occurs at the cathode. Speculateon what species is being oxidised at the anode during electrolysis?

Explanation / Answer

1) the coordination geometry of [V(OH2)6]3` and the overall structure are dictated by the electronic preference of the vanadium(III) ion leading to the formation of alums of the beta type.

2) K2Cr2O7 + 4H2SO4 + 3CH3CH2OH --> K2SO4 + Cr2(SO4)3 + 7H2O + 3CH3.CHO

Potassium dichromate (VI) solution is being oxidised at the node during electrolysis

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.